colloid, supersaturation, coagulation, peptization, counter ion, coprecipitation, occlusion,

nucleation, homogeneous precipitation, post precipitation, surface adsorption.

I+ Supersaturation curveI

CH 233 (H) 5 5

56 CH 233 (H)

CH 233 (H) 57

10 cmCD- m

1 c m

10 cm.

dUiiih 1 &UlJ0~~lJ~fllJ7fl#h~Fl10 cm3 = 10x 10 = 100 cm’1 *

58 CH 233 (H)

CH 233 (H) 59

CH 233 (H)

= 0.1909 ns’u. ~,H~nvoJ~julsln~~~~~J~u~. . = 0.5208+1.1110+0.1909 n=hJ

= 1.823 tls’oJ

= 0.2345 ns”u

w t KC103 = wtO2. gfw KC103 2gfw 02 . i

= 0.2345xyx;

= os9a7 ns’u

a ) :. %J KC103 = o.5987-x1000.8000

= 74.84

b) wt AgCl = wt KCIOl gfw AgC1gfw KC103

= 0.5987~ 143.32122.55

= 0.7002 n<aJ

64 CH 233 (H)

wt KClOx = wt AgCl .gfw KClOxgfw AgCl

0.2800 = 0.2900 x (74-;&+;;xl

74.56+16x = 138.38

x - 63.8216

xx8fw AgBrx2g f w CaBrl +“$%% = o’7500

2xx187.78+yx187.78 = 0*75tjo199.88 102.90

1.8789x+ 1.8248 y = 0.7500

1rnufi1 (1) Iu (2)

1.8789 (0.4050 - y) + 1.8248 y = 0.7500

0.7610-1.8789y+1.8248y = 0 .7500.

0.0541 y = 0.0110

y = 0 .2033 idll

. . ...(l)

. . ...(2)

CH 233 (H) 05

lltfFid1 NaBr ‘Mu”n = 0.2033 fis”u

:. CaBr2 dfl = 0.2017 i+iJ

a ) ILkl~t’hJdWl~ NaBr ~U~l5d?Fdl4

0.2033 xloo=-0.4050

= 50.20 %

‘3

wt CaBra . gfw Cao x 100

Vo CaO =gfw CaBrl

wt sample

0.2017x s*x loo=

0.4050

= 13.97%

66 CH 233 (H)

no. fw. wt NalSx = i no. fw. wt. BaS04

wt Na& 1 wt BaS04= -.gfw Na& x gfw BaSO.,

0.2ooo 1 1.0842- -(46 + 32 x) = x 233.40

43.0548x = 46+32x

11.0548x = 4 6

46x=11.o548

= 4.2

.‘. x = 4

23) F1lX-KW CaC03, CaSO, bbRt BaC03 dXfKlU~‘Xl 28.5% Ca bbWt 32.0% CO,D

odiiwaslbPslbdO$bhiiPa6YO9 CaC03, CaSO, bb~‘~ BaC03 %p68lSWsXJ~

% Ca ?ilJlQlfl CaC03wt CaCOpx gfw Ca xltxl

gfw CaCOl=wt sample

wt CaC03x$$$x 100=

wt sample

40.04 wt CaCOa.cwt sample

% Ca ?lJl9lfl CaSO.+wt CaS04x gfw Ca

gfw CaS04xl00

=wt sample

wt CaS04 x s4 X 100rz

wt sample

= 29.44wt CaS04wt sample

CH 233 (H) 67

.. . 40.04 wt CaCO3 + 29.44 wt CaS04wt sample wt sample

= 28.5

~u~~uo&r$‘u v. cot h’lrl~diiodl~~~lJlC7fl CaCOs tW BaCOJ

. . . . . ...(l)

070 co2 ?ilJlWl CaCOtwt CaC03. -44-o1 x loo

100.097wt sample

43.91 wt CaCOs=wt sample

% CO2 dIJlQlfl BaCOswt BaCO3. -44*o1 xl00

197.35twt sample

= 22.28 wt BaCOswt sample

,. . 43.97 wt CaC03+22.28 wt BaCO, = 32 owt sample wt sample

. . . . . ...(2)

wt BaCOJ = wt sample - wt CaCOt - wt CaSO4

43.97 wt CaC03 + 22.28 (wt sample - wt CaCCh - wt CaSOd

= 3.20 wt sample

21.69 wt C&O3 - 22.28 wt CaSO, = - 19.08 wt sample . . . . . ...(3)

owl (1)

40.04 wt CaCOJ + 29.44 wt CaS04 = 28.5 wt sample . . . . . ...(4)

flltl (3) 11% (4) IlVil wt CaC03 LLBZ CaSO, OZIQI’

wt CaC03 = 0.0479 wt sample

wt CaSO, = 0.9029 wt sample

wt BaCOJ = wt sample - 0.0479 wt sample - 0.9029 wt sample

= 0.0492 wt sample

:. q. CaC03 = 0.0479wtsam~le~,~wt sample

= 4.79%

q. CaSO, = 0.9029-~le~l~wt sample

= 90.29%

070 BaCOJ = 0.0492 wt sample x 1oowt sample

68

= 4.92olo

CH 233 (H)_’

wt TiOl .%Ti =

u.u.n15wnlJ0onlrd

wt TiOl . -47.90x loo5.00 = 19.90

0.0780

Vo Fe =wtFex100

U.U.iTl5WlQUl9

2.50 = wt FexlOO0.8OCO

:. wt Fe =

=

:. %~fl1$$3Qdl~ll~~ Fez03 Wod”n =

=

%A1 =.

=

2.5 x0.80001 0 0

0.0200

wt Fe gfw Fe203 x 1gfw Fe 2

0,02)0x 159.69 = r55.8 2

0.0286 flfl.l

0.0780-0.0286-0.0065

0.0429 nf¶J

wt A1103 x gfw Al x~x100gfw A1203 I

wt sample

0.0429x~6x2x100

0.8000

= 2.84 %

CH 233 (H) 69

25), 26), 27), 28) fhkIil$iaWlULQJ

wt LiCI+ wt KC1 = 0.5000 fls’u . . . . . . . . (1)

wt LiCl x gfw Liso4x! + wt KC1 xgfw ~04 x!g f w LiCl 2

= Q 6227 ns’ugfw KC1 2 ’

. . . . . ,..(a

wt LiClxEx!+wt KCIxwx!42.40 2 74.56 2

= 0.6227

1.30 wt LiCl+ 1.17 wt KCI = 0.6227 . . . . . ...(3)

Qlfl (1) wt LiCl = O.SOOO- wt KCI . . . . . ...(4)

mlufi1 (4) m9n (3)

1.30 (O.SWO - wt KCI) + 1.17 wt KC1 = 0.6227

0.6500 - 1.30 w t KC1 + 1.17 w t KC1 = 0.6227

0.13 w t KCI = 0.0273

wt KC1 = 0.2100

’. . wt LiCl = 0.2900 fls’u

Cl ihlfl KC1 = wt KClx -$$,

= o.2100x~6

= 0.0998 f&J

Cl ViIJlWfI LiCl = wt LiClx gfwgfw LiCl

= 0.2425 t-&J

:. C l Gh+hJk? = 0.0998iO.2425

= 0.3423

%CI = ;+x100

1lJfJhiudl C l Iwnls~~fldlJ = 68.46 '70

70 CH 233 (H)

wt TIS04 htilWi%u’l~ = wtTl1 x gfw TLSO., ~ !gfw TII 2

= o.I86x~,x;

= 0.1419,

5’0 TIzSO, = ‘ex 100

= 1.46 %

wt S = wt BaSO, gfwsgfw BaS04

= O.lSOx~$j

= 0.0206 nth

:. wt C,HSNO,S = wt Sx gfw C,HsNOjSgfw s

= 0.0206x~

CH233 (H) 71

100 mg NiCI* Fua

:. k=fWh&~%~ C.,H8N201 = 1 0 0 x 10-3.2 ha129.60

= ‘q0z;,;+x2x 116 fls’u

= 17.9 mJ.v¶J.

&leJJqhJlfuziuwD 5 %

7 2

_ 0.20 _ 5.84 x 10-4 hJn342.14

CH 233 (H)

’ Bl’Oaj$ BaC12. .

a) BaClt h&h 0.120 F

’ O:t:la’ BaCl~ 0.120 F. .

= 5.84x 10-.x3

5.84 x lo-‘x 3 x 10’=0.120

= 14.6

9h;oJl+hJIdlJlflliiuwo IO @Jo

llffRdl~tlJ(l~ 0.120 F BaCl, = 110x14.6100

= 16.1

b) h%lXKtWJ Bat& . 2H10 l&.&J 3.25 % (w/v)

tul

RlJ.TflJ.

tlu.cHoJ.

mJ.J¶J.

CH 233 (H)

wt Ba&+ wt SrC12 = 0.3705

wt BaCll . gfy BaCO3 + Wt SrCl gfw SrC03gfw BaC12 * . gfw SrCl2

= 0.3485

wt BaClz .197.35+ wt SrClz . E208.28 = 0.3485

0.9475 wt BaC11+0.9312 wt SrCll

mwfii (1) RJIW (3)

= 0.3485

0.9475 (0.3705 - wt SrCl2) + 0.93 12 wt %Cl~ = 0.3485

0.35 10 - 0.9475 wt SrCll + 0.93 12 wt SrCll = 0.3485

. .’ w t SrCla = ‘e

= 0.1534

f6J ........ (1)

fliil ...... ..(2)

n=h .. . . . . ..(3)

t-lcJ

:. wt BaCL = 0.3705 -0.1534

= 0.2171 nfiJ

w t BaClax gfw Bagfw BaCl*

x 100% B a =

wt sample

=0.2171 xex 100

0.37oi

= 38.6%

wt Src& x gfw x loo%Sr =

gfw SrClswt sample

=0.1534xE2x 100

0.370.5

= 22.89%

74 CH 233 (H)

wt NaCl = wt AgClxg rd

= 0.6655 f-ls”u

:. wt (NH&S04 = 0.7970 - 0.6625

= 0.1315 fls’u

% s o , =0.1315x=4x loo

0.7970

= 9.997

% N H , =0.1315x~4x100

0.7970

= 4.245

0.6655~258.44

xl00%Na =

c.7970

= 32.86

wt Al(C,H,ON),xgfw Al203

x~xloo

% Al203 =gfw Al (C9H60N)3 2

wt sample

0.0304x101.96.~.100458.98 2=0.7500

= 0.450

CH 233 (H) 75

wt A&l + wt AgBr = O.EooO . . . . . ...(I)

wt AgCl + wt AgBr x gfwgfw AgBr

= 0.8000-0.1066

143.32wt AgCl + wt AgBr . -187.78

= 0.6934

wt AgCl+O.7632 wt AgBr = 0.6934

Bin (1) mun~lu (2)

(0.8000 - wt AgBr) + 0.7632 wt AgBr = 0.6934

0.2368 wt AgBr = 0.1066

:. wt AgBr = 0.4502 fl55.I

’ w t AgCi. . = 0.8000 - 0.4502

= 0.3498 nTl.J

. . . . . . . . (2)

NHEONH, + H# *2NH, + Co2

3 2HaO

2NH;+20H-

Fe”+3 OH- * Fe(OH)s

F e ” dlhd = O.l@J fl’f%l 0.100= -55.85

= 1.79x lo-’ ha

olnnUnlrlmA~?'l~a~~~[ OH-] = 3x1,79x10-’ ‘IUZI

76 CH 233 (H)

%KI =wt Ba (IO& gfw KI X~Xloo

gfw Ba (IO& 1wt sample

0.0720xE4x2x I00

2.72

1.80

I .80 ?h

CH 233 (H) 7 7

1)2)

3 )

4 )

5 )

6)

7 )

8)

9 )

10)

11)

12)

13)

14)

70

What is the von Wiemarn ratio 7 Define the terms in it.

What information concerning optimum conditions of precipitation does the von Weimarnratio give us 7

What is digestion of a precipitate and why is it mecessary ?Outline the optimum conditions for precipitation that will obtain a pure and filterableprecipitate.What is coprecipitation ‘! List the different types of coprecipitatation and indicate howthey may be minimized or treated far.Why must a filtered precipitate be washed ?

A 0.6834-g sample containing chloride is dissolved and the chloride precipitated as AgCI.The precipitate is washed, dried, and found to weigh 0.4281 g. Calculate the percentagechloride in the sample.The lead in a 0.5250-g sample of an ore is precipitated as PbS04. The dried precipitateweighs 0.4264 g. Calculate (a) the percentage of lead(Pb) in the ore and (b) the percentageexpressed PbsOd.

The phosphorus in a sample of phosphate rock weighing 0.5428 g is precipitated as MgNHdP04.6HsO and ignited to MgsPsOr. If the ignited precipitate weighs 0.2234 g, calculate (a) thepercentage of PsOs in the rock and (b) the weight of the precipitate of MgNHdPOd .6HzOIron is determined in a sample by precipitating Fe(OH)a and igniting the precipitate to

FesOs. What weight of sample should be taken for analysis so that each milligram of Fe203represents 0.100% Fe in the sample?A pure sample of sodium chloride weighing 0.653 10 g is dissolved in water and the chlorideprecipitated as AgCI. If the AgCl precipitate weighs 1.6029 g, calculate the atomic weightof sodium. Assume the atomic weights of chlorine and silver are 35.453 and 107.87, respectively.A mixture which contains only Fe203 and Al203 weighs 0.6200 g. It is treated with Hz,reducing the Fe203 to Fe. The A1203 is not changed. The mixture now weighs 0.5321 g.Calculate the percentage of Al in the sample.

A sample containing only CaCO, and MgCO, is ignited to CaO and MgO. The mixture ofoxide weighs exactly 50.00% as mush as the original sample. Calculate the percentages ofCaCO3 and MgCOs in the sample.A 0.3531-g sample containing only CaCOs and MgCOs produced 89.6 ml of COz at STPwhen treated with HCI. Assuming that all the carbonates reacted completely, how manygrams of CaCOs did the original sample contain?

CH 233 (H)

15) A certain material contains 20.0% KCI. How many grams of pure NaCl should be addedto 1.40 g of this material so that the resulting mixture will contain 15.0% chloride (Cl) ?

1 6 ) The sodium and potassium in a sample weighing 0.9134 g. are converted into NaCI andKCl. The mixture of chlorides weighs05924 g.The chlorides are then treated with sulfuricacid, converting them to NazS04 and KzSO4. The mixture of Sulfates weighs 0.7024 g.

Calculate the percentages of NasO and KzO in the sample.1 7 ) A 0.6000-g sample that contains both NaCl and NaBr gives a precipitate of AgCl and

AgBr that weighs 0.4482 g. Another 0.6000 g sample is titrate, requiring 26.48 ml of0.1084-M AgNOs for complete reaction. Calculate the percentages’of NaCl and NaBrin the sample.

1 8 ) A mixture containing only AgCl and AgBr weighs 0.4834 g. It is treated with Clz, convertingthe AgBr to AgCI. The total weight of AgCl is 0.3826 g. Calculate the weight of AgBr in

the original sample.

CH 233 (H) 79